CASH-Editorial

In this problem there are n bags and each bag contain some coins and chef wants that each bag contain coins divisible by k,
so,the idea is we took money from bag which have some extra money than needed means that let a bag contain coins x,and x is not divisible by k so we take out extra coins and make it divisible to k.
After taking extra coins from every bag we just put them in last bag.
And now take extra coin from last bag.
Sample

for(int i=0;i<n;i++)
{
cin>>a[i];
if(a[i]%k!=0)
{sum = sum + (a[i]%k);}
}

now tkae mod of sum with k and you get the answer.

Code for this problem

/*hard cash*/

/****killing like kamikaze****/
                      
     /*tanuj yadav*/                

/**********************/

#include<bits/stdc++.h>
using namespace std;

int main()
{
int t,n,k;
cin>>t;
for(int m=0;m<t;m++)
{
cin>>n>>k;
long int a[n];
long long int sum =0;
for(int i=0;i<n;i++)
{
cin>>a[i];
if(a[i]%k!=0)
{sum = sum + (a[i]%k);}
}	
cout<<sum%k<<endl;
}
	return 0;
}

but how can you say this is going to be correct answer. what if a combination where the c part hinders

1 Like

Sorry i cant understand your query can you elaborate what do you mean by c.

but your code will give 1 answer instead of 0 when bag contains: 1,13 and 1 coins